That looks correct - you can always test your expressions - if the result changes you know you've done it wrong.
Personally, though, it's difficult for me to imagine how an inverse of a ;._3 adverb would work. Logically, I suppose, you might be looking for something that tessellates the result (which we usually do by producing a result with a couple extra dimensions and then combine them using ,/)? Thanks, -- Raul On Sat, Dec 2, 2017 at 1:02 PM, Andrew Dabrowski <[email protected]> wrote: > On 12/02/2017 02:17 AM, Roger Hui wrote: > >>>> SC =: 3 : '(3 3$4>i.5) ,./^:2@(*/)^:y ,.1' >>> >>> SC confuses me. I would have thought that >>> >>> (3 3$4>i.5) (*/)^:y ,.1 >> >> The left operand of the power operator ^: is ,./^:2@(*/) . >> > Oh, I parsed it as > > (3 3$4>i.5) (,./^:2) @ ((*/)^:y) ,.1 > > but it should be > > (3 3$4>i.5) (,./^:2 @ */)^:y ,.1 > > is that right? I'm certainly not au courant in all the exceptions to > right-association. > > Btw, does the cut function ;._3 have an inverse that could be used to solve > this problem? > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
